How to show $k[x]$ is not isomorphic to $k[x,y]/(xy+x^3+y^3)$?

In your ring, $(x + y)^3 = xy(3x + 3y - 1)$.

The ring $k[x]$ does not have any element $g$ such that $3g - 1$ divides $g^3$. This follows by factoring $g$ into linear factors, for example.

The idea you already had is great. The localization of your ring at the ideal $(x, y)$ is not regular, by the Jacobi criterion. The localization of $k[x]$ at any ideal is regular.

Here is an elementary solution. Let $f : k[x,y]/(xy+x^3+y^3) \to k[z]$ be any ring homomorphism (using another variable for the codomain to avoid confusion). Let $p = f(x)$ and $q = f(y)$ so that we know $pq + p^3 + q^3 = 0$. Now, let $r$ be a root of $p$. We see that $$0 = 0(r) = (pq + p^3 + q^3)(r) = 0 q(r) + 0^3 + q(r)^3 = q(r)^3,$$ so $r$ is also a root of $q$. Symmetrically, every root of $q$ is also a root of $p$, so $p$ and $q$ have the same set of roots. Since $k$ is algebraically closed, this means that $p$ and $q$ are associates in $k[z]$, i.e. $p = \alpha q$ for some $\alpha \in k \setminus \{0\}$. But now $pq + p^3 + q^3 = 0$ tells us that $$\alpha p^2 + p^3 + \alpha^3 p^3 = 0,$$ which implies that $p = 0$ or $\alpha + (1+\alpha)p = 0$. In the latter case, we see that $\alpha \neq -1$, and therefore $p = \frac{-\alpha}{1+\alpha}$. Thus, in either case, we have $p \in k$, so also $q = \alpha p \in k$. But now $f$ cannot be surjective, since the image of $f$ is the subring of $k[z]$ generated by $k \cup \{f(x), f(y)\} = k \cup \{p,q\} = k$, which is just $k$.

This shows the slightly stronger fact that no quotient of $k[x,y]/(xy+x^3+y^3)$ is isomorphic to $k[x]$.

Edit: I made a very silly mistake in the above "proof" – having the the same set of roots does not make two polynomials in $k[z]$ associates! They need the same roots with the same multiplicities. Maybe the above argument can be modified to fix this issue, but I don't see it immediately and I'm going to bed now. I'll leave this up for posterity but please don't use this answer to convince your friend!